Two insulated charged spheres of radii `20 cm` and `25 cm` respectively and having an equal charge `Q` are connected by a copper wire, they are separated

Options

A. Both the spheres will have the same charge `Q`
B. Charge on the `20 cm` sphere will be greater than that on the `25 cm` sphere
C. Charge on the `25 cm` sphere will be greater than that on the `20 cm` sphere
D. Charge on each of the sphere will be `2Q`

Explanation

To solve the problem step by step, we need to analyze the situation involving the two charged spheres connected by a copper wire. Here’s how we can approach the solution:

### Step 1: Understand the Initial Conditions
We have two insulated charged spheres:
- Sphere 1 with radius \( R_1 = 20 \, \text{cm} = 0.2 \, \text{m} \)
- Sphere 2 with radius \( R_2 = 25 \, \text{cm} = 0.25 \, \text{m} \)

Both spheres have an equal charge \( Q \).

### Step 2: Connect the Spheres
When the spheres are connected by a copper wire, charge can flow between them until they reach the same electric potential. The electric potential \( V \) of a charged sphere is given by:
\[ V = \frac{kQ}{R} \]
where \( k \) is Coulomb's constant.

### Step 3: Set Up the Equation for Electric Potential
Since the potentials of both spheres must be equal when connected:
\[ V_1 = V_2 \]
This gives us:
\[ \frac{kQ_1}{R_1} = \frac{kQ_2}{R_2} \]
Cancelling \( k \) from both sides, we have:
\[ \frac{Q_1}{R_1} = \frac{Q_2}{R_2} \]

### Step 4: Substitute the Radii
Substituting the values of \( R_1 \) and \( R_2 \):
\[ \frac{Q_1}{0.2} = \frac{Q_2}{0.25} \]

### Step 5: Express \( Q_1 \) in Terms of \( Q_2 \)
Cross-multiplying gives:
\[ Q_1 \cdot 0.25 = Q_2 \cdot 0.2 \]
Thus,
\[ Q_1 = \frac{0.2}{0.25} Q_2 = \frac{4}{5} Q_2 \]

### Step 6: Total Charge Conservation
The total charge before connecting the spheres is:
\[ Q_1 + Q_2 = 2Q \]
Substituting \( Q_1 \) from the previous step:
\[ \frac{4}{5} Q_2 + Q_2 = 2Q \]
Combining the terms:
\[ \frac{4}{5} Q_2 + \frac{5}{5} Q_2 = 2Q \]
\[ \frac{9}{5} Q_2 = 2Q \]

### Step 7: Solve for \( Q_2 \)
Multiplying both sides by \( \frac{5}{9} \):
\[ Q_2 = \frac{10}{9} Q \]

### Step 8: Find \( Q_1 \)
Using \( Q_1 = \frac{4}{5} Q_2 \):
\[ Q_1 = \frac{4}{5} \cdot \frac{10}{9} Q = \frac{8}{9} Q \]

### Step 9: Conclusion
Now we have:
- Charge on Sphere 1 (20 cm radius): \( Q_1 = \frac{8}{9} Q \)
- Charge on Sphere 2 (25 cm radius): \( Q_2 = \frac{10}{9} Q \)

Since \( Q_2 > Q_1 \), the charge on the sphere with a radius of 25 cm is greater than that on the sphere with a radius of 20 cm.

### Final Answer
The charge on the sphere with radius 25 cm is greater than the charge on the sphere with radius 20 cm.